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\(\sum_{n=2}^{\infty} \frac{(-1)^n}{n\log{n}}\)

Calculus (Integrals, Series)
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Grigorios Kostakos
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\(\sum_{n=2}^{\infty} \frac{(-1)^n}{n\log{n}}\)

#1

Post by Grigorios Kostakos » Sat Mar 04, 2017 1:12 pm

Evaluate the convergent series \[\mathop{\sum}\limits_{n=2}^{\infty} \frac{(-1)^n}{n\log{n}}\,.\]


Note: I don't have a solution for this.
Grigorios Kostakos
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Re: \(\sum_{n=2}^{\infty} \frac{(-1)^n}{n\log{n}}\)

#2

Post by Riemann » Sat Dec 14, 2019 3:16 pm

Basically it equals to

$$\int_1^\infty \left( 1+\left(2^{1-s}-1\right)\zeta(s) \right) \, \mathrm{d}s$$

However, the $\zeta$ function does not behave well under integrals. So, I would not expect a closed form to exist ... !
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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